Lecture 4: Normalized Functions of Bounded Variation; Regular Borel Measures; the Riesz–Markov–Kakutani Theorem;

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Lecture 4: Normalized Functions of Bounded Variation; Regular Borel Measures; the Riesz–Markov–Kakutani Theorem; Functional analysis Lecture 4: Normalized functions of bounded variation; regular Borel measures; the Riesz{Markov{Kakutani theorem; Remark 3.10. Assume that the functional Λ above is positive (i.e. Λf ≥ 0 for f ≥ 0). We observe that the extension Λe is also positive and hence the constructed function f is then monotone increasing. Indeed, suppose that there is g 2 `1[0; 1] with g(t) ≥ 0 for each t 2 [0; 1] but Λeg < 0. By normalizing, we can assume that kgk1 = 1, thus 0 ≤ g(t) ≤ 1 for each t 2 [0; 1]. Let 1 be the constant 1 function. As Λ is positive and linear, it is monotone (i.e. Λg ≤ Λh if g ≤ h), and hence Λ1 = kΛk. We have k1 − gk1 ≤ 1 and Λ(e 1 − g) = Λ1 − Λeg = kΛk − Λeg > kΛk = kΛek; which gives a contradicition. Therefore, positive continuous functionals on C[a; b] correspond to increasing func- tions on [a; b], while general continuous functionals correspond to functions of bounded variation. Remark 3.11. Since every f 2 BV([a; b]) is the difference of two increasing functions, we see that every Λ 2 (C[a; b])∗ can be written as Λ = Λ+ − Λ−; where Λ+; Λ− are positive. (3.1) The conclusion of Remark 3.11 can be justified without appealing to Riesz' Theo- rem 3.9. To this end, let us use a more general language which will be also useful in the sequel. By an ordered vector space we mean any vector space E over R equipped with a partial order relation ≥ compatible with the algebraic operations in the sense that x ≥ y implies x + z ≥ y + z for all x; y; z 2 E and x ≥ y implies λx ≥ λy for all x; y 2 E, λ ≥ 0. We denote by E+ the positive cone of E defined by fx 2 E : x ≥ 0g. An ordered vector space which is also a lattice with respect to the given order is called a Riesz space or a vector lattice. If E is a Riesz space and _, ^ denote the lattice operations, then for any vector x 2 E we define its positive part x+, negative part x− and its modulus jxj by x+ = x _ 0; x− = (−x) _ 0; jxj = x _ (−x): We then have x = x+ − x− and jxj = x+ + x−, so this gives a decomposition of x as the difference between two positive elements. A Riesz space is called Archimedean if whenever 0 ≤ nx ≤ y for all n 2 N and some y 2 E+, then x = 0. If E is a Riesz space and a normed space with a norm k·k satisfying the condition jxj ≤ jyj =) kxk ≤ kyk for all x; y 2 E; then E is called a normed Riesz space, and in the case where the norm is complete, we call E a Banach lattice. There are plenty of examples of Banach lattices: Rn with the pointwise ordering and any of the norms k·kp (1 ≤ p ≤ 1), the space C(X) of continuous 1 functions, and Cb(X) of bounded continuous functions on any topological space X, also considered with the pointwise ordering and the supremum norm, as well as Lp(µ)-spaces under the a.e. pointwise ordering. If E is a normed Riesz space, then its dual E∗ can be ordered in a natural way by defining f ≥ g if and only if f(x) ≥ g(x) for every x 2 E+ (f; g 2 E∗). We will see (Lemma 3.18 below) that in this way we equip E∗ with a Banach lattice structure1. In particular, the decomposition (3.1) is possible in the Banach lattice (C[a; b])∗. Actually, decomposition (3.1) holds true with a further requirement that kΛk = kΛ+k + kΛ−k and in a much more general setting of the so-called order unit normed Riesz spaces (see Problem 3.22). For C[a; b] we will obtain that equality with the aid of Hahn's decom- position theorem. As we know that every continuous linear functional on the real space C[a; b] is rep- resented by a function f 2 BV[a; b], let us consider the problem of uniqueness of this representation. Obviously, f is not unique as stated in Theorem 3.9, because adding any constant function to f does not change the Riemann{Stieltjes integral. So, one normalization would be to require that e.g. f(a) = 0. By Jordan's decomposition, we have f = g −h for some increasing functions g and h, and therefore f has at most countably many points of discontinuity in (a; b). Observe that by modifying the values of f at each of these points, but keeping the variation bounded, we do not change the value of the integral in formula (3.7). Indeed, any g 2 C[a; b] is Riemann{Stieltjes integrable, hence its integral can be approximated by any sequence of integral sums over partitions with diameters converging to zero2. In particular, we can avoid each point of discontinuity of f (except the endpoints) in these partitions. Consequently, another normalization condition which does not affect the validity of (3.7) is that f(t+) = f(t) for each t 2 (a; b), i.e. f is right- continuous inside the interval (a; b) (the choice of the left or right limit is arbitrary). Every f 2 BV([a; b]) satisfying the above two normalization conditions is called a normalized function of bounded variation, and we denote the class of all such functions by • NBV([a; b]) = ff 2 BV([a; b]): f(a) = 0 and f is right-continuous in (a; b)g. By modifying the representing function f in Theorem 3.9 as described above, we can guarantee that it belongs to the class NBV([a; b]). Moreover, it turns out that in this class the function f is unique. We will obtain this as a particular case of the `uniqueness part' of Theorem 3.16 below. Before proceeding to that result, let us build a `bridge' be- tween the special Riesz Theorem 3.9 and the much more general Riesz{Markov{Kakutani Theorem 3.16. Given any function f 2 BV([a; b]), we define a measure µf on the field F consisting of finite unions of the intervals jy; x](a ≤ y < x ≤ b) which are all right-closed and left-open unless y = a in which case we take the closed interval [a; x]. Namely: • µf ([a; x]) = f(x) − f(a), • µf ((y; x]) = f(x) − f(y) for a < y < x ≤ b. 1An interested reader may consult e.g. [C.D. Aliprantis, K.C. Border, Infinite dimensional analysis. A hitchhiker's guide, 3rd edition, Springer 2006; Ch. 8 and 9]. 2See [W Rudin, Principles of mathematical analysis; Thms. 6.6 and 6.8]. 2 Notice that the measure µf is finitely additive and satisfies n n X [ µf (E) = (f(bj) − f(aj)) for E = ja1; b1] [ (aj; bj] 2 F; j=1 j=2 where a ≤ a1 < b1 ≤ a2 < b2 ≤ ::: ≤ an < bn ≤ b. Moreover, µf is bounded because f is of bounded variation. But, as we will see, µf has also another nice property of being regular, provided that f belongs to the class NBV([a; b]). Definition 3.12. Let F be a field of subsets of a topological space X and µ: F! K be a finitely additive set function, where K 2 fR; Cg. We say µ is regular if for every E 2 F and " > 0 there exist a set F 2 F with F ⊆ E and a set G 2 F with int G ⊇ E satisfying jµ(C)j < " for every C 2 F;C ⊆ G n F: If f 2 NBV([a; b]), it is possible to extend µf to a regular σ-additive measure defined on all Borel subsets of [a; b]. It follows from the following general result on extensions of measures. (Actually, we only need a simplified version of this result, as the measure µf is trivially σ-additive in the sense that there are no countably infinite, but not finite, unions of the intervals jc; d] in F.) Extension theorem. Let F be a field of subsets of some topological space X and let µ: F! K be a bounded, regular, finitely additive set function, where K 2 fR; Cg. Then, µ has a unique regular, σ-additive extension µe: σ(Σ) ! K to the σ-algebra σ(F) generated by F. For the proof see e.g. [N. Dunford, J.T. Schwartz, Linear operators (vol. 1: General theory), Wiley{Interscience 1988; Ch. III.5]. This theorem is a combination of two classical measure theory results: • the Alexandroff theorem which says that every bounded regular finitely additive (real- or complex-valued) set function on a field of subsets of a compact topological space is σ-additive; • the Carath´eodory extension theorem which guarantees that positive σ-additive mea- sures defined on a ring R can be extended to positive σ-additive measures on the σ-algebra generated by R. (It should also be noted that a measure is regular if and only the positive and negative parts of its real and imaginary parts are regular.) Lemma 3.13. If f 2 BV([a; b]), then for every t 2 [a; b) we have lim V (f; [t; t + "]) = 0: "!0+ Proof. (classes) Now, assume that f 2 NBV([a; b]). Since f is right-continuous on (a; b), we have lim"!0 f(t + j"j) = f(t) for every t 2 (a; b) and it follows that for every interval jc; d] 2 F, n n X Sn o V (f; [c; d]) = sup jµf (Ij)j: Ij 2 F;Ij \ Ik = ? for j 6= k; j=1Ij ⊆ jc; d] : j=1 3 (The right-hand side defines the variation of µf on the set jc; d]; we will introduce this notion later on.) Plainly, we can extend this formula to any set E 2 F in the sense that if E = I1 [ ::: [ Im, with Ij = jcj; dj] 2 F (1 ≤ j ≤ m), then the left-hand side is Pm j=1 V (f; [aj; bj]), whereas the right-hand side is the variation of µf on E, denoted by jµf j(E).
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